Question: Let $f(x)=2x^2-\dfrac{1}{x^2}+\dfrac{1}{x^3}$. $f'(x)=$
Answer: The strategy We can first rewrite each rational term of $f$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} f(x)&=2x^2-\dfrac{1}{x^2}+\dfrac{1}{x^3} \\\\ &=2x^2-x^{-2}+x^{-3} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(2x^2-x^{-2}+x^{-3}) \\\\ &=2\dfrac{d}{dx}(x^2)-\dfrac{d}{dx}(x^{-2})+\dfrac{d}{dx}(x^{-3}) \\\\ &=2( 2x)-(-2)x^{-3}+(-3)x^{-4} \\\\ &=4x+\dfrac{2}{x^3}-\dfrac{3}{x^4} \end{aligned}$ In conclusion, $f'(x)=4x+\dfrac{2}{x^3}-\dfrac{3}{x^4}$.